Show that any graph having five or fewer vertices and a vertex of degree 2 is planar.

I don't understand the question.

Are they saying 5 or fewer vertices AND 1 vertex with a degree of 2?
5 or fewer vertices with degrees of 2?

Is it asking something else?

If the first, could I put 3 vertices and one with a degree of 2?
If the second, is that possible?

Lastly, beyond drawing a graph, is there any way to show this?

The answer is the back of the book is as follows:
"a graph of five or fewer vertices and a vertex of degree 2 is homeomorphic to a graph with four or fewer vertices. Such a graph cannot contain a homeomorphic copy of "

Thanks!

Thanks!

Oct 25th 2010, 10:18 PM

emakarov

Quote:

Are they saying 5 or fewer vertices AND 1 vertex with a degree of 2?

Yes, I believe the assumption is saying that a graph has at most 5 vertices, one of which has degree 2.

Let v be the vertex of degree 2, and let v1, v2 be the vertices it is connected to. The hint makes sense to me provided v1 and v2 are not connected: then v can be eliminated and an edge (v1, v2) can be added to create a homeomorphic graph. I am not sure how to proceed with the proof if v1, v2 are connected.